Question: Let $g$ be a differentiable function with $g(5)=8$ and $g'(5)=0$. What is the value of the approximation of $g(5.1)$ using the function's local linear approximation at $x=5$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $8$ (Choice B) B $8.1$ (Choice C) C $8.2$ (Choice D) D $8.3$
The local linear approximation of $g$ at $x=5$ is achieved using the equation of the line tangent to $g$ at $x=5$. Let $L(x)$ represent this equation. We can find $L(x)$ using the general formula for the tangent to the graph of function $u$ at $x=a$ : $y=u'(a)(x-a)+u(a)$ [Is there a way to find this formula without memorizing?] In our case, $L(x)=g'(5)(x-5)+g(5)$. Plugging $g(5)=8$ and $g'(5)=0$, we obtain $L(x)=0(x-5)+8=8$. To approximate $g(5.1)$, all we need is to plug $x=5.1$ into $L(x)$. $\begin{aligned} L(5.1)&=8 \end{aligned}$ In conclusion, the approximation of $g(5.1)$ using the function's local linear approximation at $x=5$ is $8$.